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omeli [17]
3 years ago
11

A. Use average bond energies together with the standard enthalpy of formation of C(g) (718.4 kJ/mol ) to estimate the standard e

nthalpy of formation of gaseous benzene, C6H6(g). (Remember that average bond energies apply to the gas phase only.)
B. Compare the value you obtain using average bond energies to the actual standard enthalpy of formation of gaseous benzene, 82.9 kJ/mol. What does the difference between these two values tell you about the stability of benzene?

Chemistry
1 answer:
Nana76 [90]3 years ago
5 0

Answer:

a. 278.4kJ/mol

b. due to resonance, some energy are released

Explanation:

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In the laboratory a student combines 40.6 mL of a 0.113 M copper(II) sulfate solution with 26.4 mL of a 0.329 M copper(II) iodid
Arisa [49]

Answer : The final concentration of copper(II) ion is, 0.198 M

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First we have to calculate the moles of CuSO_4 and CuI_2.

\text{Moles of }CuSO_4=\text{Concentration of }CuSO_4\times \text{Volume of solution}

\text{Moles of }CuSO_4=0.113mol/L\times 0.0406L=0.00459mol

Moles of CuSO_4 = Moles of Cu^{2+} = 0.00459 mol

and,

\text{Moles of }CuI_2=\text{Concentration of }CuI_2\times \text{Volume of solution}

\text{Moles of }CuI_2=0.329mol/L\times 0.0264L=0.00869mol

Moles of CuI_2 = Moles of Cu^{2+} = 0.00869 mol

Now we have to calculate the total moles of copper(II) ion and total volume of solution.

Total moles copper(II) ion = 0.00459 mol + 0.00869 mol

Total moles copper(II) ion = 0.0133 mol

and,

Total volume of solution = 40.6 mL + 26.4 mL = 64.0 mL = 0.067 L    (1 L = 1000 mL)

Now we have to calculate the final concentration of copper(II) ion.

\text{Final concentration of copper(II) ion}=\frac{\text{Total moles}}{\text{Total volume}}

\text{Final concentration of copper(II) ion}=\frac{0.0133mol}{0.067L}=0.198mol/L=0.198M

Thus, the final concentration of copper(II) ion is, 0.198 M

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15.7 g / 79.9 g = 0.196 moles of atoms
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